Question

The area of a rectangle is `20` `cm^2`. The length is `8` more than the width. How do you find the width in cm?

Answer 1

The width of the rectangle is `2` cm.

Explanation:

Let's first recall the formula for the area of a rectangle:

`A=lw`

We can first substitute `20` into the area part of the equation since we already know its value:

`20=lw`

Now what? We have two variables but how do we find either of them? Looking back at the question, we see that "the length is 8 more than the width." Voila! Now we can write the length in terms of the width! So now we can switch out `l` for `w+8`:

`20=(w+8)w`

Now, using the Distributive Property, we can expand out the right side of the equation:

`20=w^2+8w`

We can subtract `20` from both sides to obtain a classic quadratic equation:

`w^2+8w-20=0`

Now, we can use factoring to simplify this problem. So our task now is to find two factors of `-20` such that they add up to `8`. We will list out all the factors of `-20` and then mark out the pair that adds up to `8` (I will use red for the pairs that don't work and blue for the pair that does work):

`color(red)(-1 and 20)`

`color(red)(1 and -20)`

`color(red)(-4 and 5)`

`color(red)(4 and -5)`

`color(blue)(-2 and 10)`

`color(blue)(2 and -10)`

We found our pair! 2 and -10 or -2 and 10! So now, we can write the quadratic equation in an `(w+a)(w+b)` format and make it equal to `0`:

`(w-2)(w+10)=0`

And since the product of `(w-2)` and `(w+10)` is 0, one of them must be equal to `0`. Let's start by assuming that `(w-2)` is equal to `0` and solve for `w`:

`(w-2)=0`
`w=2`

Now let's assume that `(w+10)` is equal to `0` and solve for `w`:

`(w+10)=0`
`w=-10`

Now, we can see that `w` could be either `-10` or `2`. However, there is no such thing as negative length so the possibility that `w=-10` is disqualified and `w=2` is the answer.